Improve knowledge base

1 files changed, 118 insertions, 147 deletions

diff --git a/source/know/concept/jellium/index.md b/source/know/concept/jellium/index.md
index 5c50f80..5cd8483 100644
--- a/source/know/concept/jellium/index.md
+++ b/source/know/concept/jellium/index.md
@@ -12,32 +12,24 @@ layout: "concept"
 **Jellium**, also called the **uniform** or **homogeneous electron gas**,
 is a theoretical material where all electrons are free,
 and the ions' positive charge is smeared into a uniform background "jelly".
-This simple model lets us study electron interactions easily.
+This is a version of the [Fermi gas](/know/concept/fermi-gas/) model,
+which we extend by including electron-electron interactions using
+[time-independent perturbation theory](/know/concept/time-independent-perturbation-theory/).
 
 
-## Without interactions
 
-Let us start by neglecting electron-electron interactions.
-This is clearly a dubious assumption, but we will stick with it for now.
-For an infinitely large sample of jellium,
-the single-electron states are simply plane waves.
-We consider an arbitrary cube of volume $$V$$,
-and impose periodic boundary conditions on it,
-such that the single-particle orbitals are (suppressing spin):
+## 0th order
 
-$$\begin{aligned}
-    \Inprod{\vb{r}}{\psi_{\vb{k}}}
-    = \psi_{\vb{k}}(\vb{r})
-    = \frac{1}{\sqrt{V}} \exp(i \vb{k} \cdot \vb{r})
-    \qquad \quad
-    \vb{k} = \frac{2 \pi}{V^{1/3}} (n_x, n_y, n_z)
-\end{aligned}$$
+Let us start with the 0th order of the perturbation expansion.
+Without interactions or potentials, this is simply a Fermi gas,
+so the single-electron wavefunctions are just plane waves.
+For mathematical convenience, we consider these waves
+in a cube of volume $$V$$ with periodic boundaries,
+leading to a discrete spectrum of allowed wavevectors $$\vb{k}$$,
+which becomes continuous for $$V \to \infty$$.
 
-Where $$n_x, n_y, n_z \in \mathbb{Z}$$.
-This is a discrete (but infinite) set of independent orbitals,
-so it is natural to use the
-[second quantization](/know/concept/second-quantization/)
-to write the non-interacting Hamiltonian $$\hat{H}_0$$,
+The unperturbed many-particle Hamiltonian $$\hat{H}_0$$ is given below in the
+[second quantization](/know/concept/second-quantization/),
 where $$\hbar^2 |\vb{k}|^2 / (2 m)$$ is the kinetic energy
 of the orbital with wavevector $$\vb{k}$$, and $$s$$ is the spin:
 
@@ -46,10 +38,9 @@ $$\begin{aligned}
     = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}
 \end{aligned}$$
 
-Assuming that the temperature $$T = 0$$,
-the $$N$$-electron ground state of this Hamiltonian
-is known as the **Fermi sea** or **Fermi sphere** $$\Ket{\mathrm{FS}}$$,
-and is constructed by filling up the single-electron states
+Which, at absolute zero $$T = 0$$, has an $$N$$-electron ground state 
+known as the *Fermi sphere* $$\Ket{\mathrm{FS}}$$
+that is constructed by filling up the single-electron states
 starting from the lowest energy:
 
 $$\begin{aligned}
@@ -57,63 +48,9 @@ $$\begin{aligned}
     = \prod_{s} \prod_{j = 1}^{N/2} \hat{c}_{s,\vb{k}_j}^\dagger \Ket{0}
 \end{aligned}$$
 
-Because $$T = 0$$, all the electrons stay in their assigned state.
-The energy and wavenumber $$|\vb{k}|$$ of the highest filled orbital
-are called the **Fermi energy** $$\epsilon_F$$ and **Fermi wavenumber** $$k_F$$,
-and obey the expected kinetic energy relation:
-
-$$\begin{aligned}
-    \boxed{
-        \epsilon_F
-        = \frac{\hbar^2}{2 m} k_F^2
-    }
-\end{aligned}$$
-
-The Fermi sea can be visualized in $$\vb{k}$$-space as a sphere with radius $$k_F$$.
-Because $$\vb{k}$$ is discrete, the sphere's surface is not smooth,
-but in the limit $$V \to \infty$$ it becomes perfect.
-
-Now, we would like a relation between the system's parameters,
-e.g. $$N$$ and $$V$$, and the resulting values of $$\epsilon_F$$ or $$k_F$$.
-The total population $$N$$ must be given by:
-
-$$\begin{aligned}
-    N
-    = \sum_{s} \sum_{\vb{k}} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}
-    = \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} \dd{\vb{k}}
-\end{aligned}$$
-
-Where we have turned the sum over $$\vb{k}$$ into an integral with a constant factor,
-by using that each orbital exclusively occupies a volume $$(2 \pi)^3 / V$$ in $$\vb{k}$$-space.
-
-At zero temperature, this inner product can only be $$0$$ or $$1$$,
-depending on whether $$\vb{k}$$ is outside or inside the Fermi sphere.
-We can therefore rewrite using a
-[Heaviside step function](/know/concept/heaviside-step-function/):
-
-$$\begin{aligned}
-    N
-    = \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \Theta(k_F - |\vb{k}|) \dd{\vb{k}}
-    = 2 \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \Theta(k_F - |\vb{k}|) \dd{\vb{k}}
-\end{aligned}$$
-
-Where we realized that spin does not matter,
-and replaced the sum over $$s$$ by a factor $$2$$.
-In order to evaluate this 3D integral,
-we go to [spherical coordinates](/know/concept/spherical-coordinates/)
-$$(|\vb{k}|, \theta, \varphi)$$:
-
-$$\begin{aligned}
-    N
-    &= \frac{V}{4 \pi^3} \int_0^{2 \pi} \int_0^\pi \int_0^\infty \Theta(k_F - |\vb{k}|) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi}
-    \\
-    &= \frac{V}{4 \pi^3} 4 \pi \int_0^{k_F} |\vb{k}|^2 \dd{|\vb{k}|}
-    = \frac{V}{\pi^2} \bigg[ \frac{|\vb{k}|^3}{3} \bigg]_0^{k_F}
-    = \frac{V}{3 \pi^2} k_F^3
-\end{aligned}$$
-
-Using that the electron density $$n = N/V$$,
-we thus arrive at the following relation:
+From our analysis of the Fermi gas, we have an important result
+for the wavenumber $$k_F = |\vb{k}_{N/2}|$$ of the highest filled orbital,
+as a function of the particle density $$n = N / V$$:
 
 $$\begin{aligned}
     \boxed{
@@ -122,103 +59,137 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-This result also justifies our assumption that $$T = 0$$:
-we can accurately calculate the density $$n$$ for many conducting materials,
-and this relation then gives $$k_F$$ and $$\epsilon_F$$.
-It turns out that $$\epsilon_F$$ is usually very large
-compared to the thermal energy $$k_B T$$ at reasonable temperatures,
-so we can conclude that thermal fluctuations are negligible.
-
-Now, $$\epsilon_F$$ is the highest single-electron energy,
-but about the total $$N$$-particle energy $$E^{(0)}$$?
+Now, let us calculate the total $$N$$-particle ground state
+energy $$E^{(0)}$$ of the unperturbed system:
 
 $$\begin{aligned}
     E^{(0)}
     = \matrixel{\mathrm{FS}}{\hat{H}_0}{\mathrm{FS}}
-    = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}
+    = 2 \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}
 \end{aligned}$$
 
-Once again, we turn the sum over $$\vb{k}$$ into an integral,
-and recognize the spin's irrelevance:
+Where we have recognized the spin's irrelevance,
+by replacing the sum over $$s$$ with a factor $$2$$.
+Next, we turn the sum over the allowed $$\vb{k}$$-values into an integral,
+which is a [common trick](/know/concept/discrete-spectrum-summation/)
+enabled by our periodic boundary conditions, yielding:
 
 $$\begin{aligned}
     E^{(0)}
-    &= \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \frac{\hbar^2 |\vb{k}|^2}{2 m}
+    &= \frac{2 V}{(2 \pi)^3} \int_{-\infty}^\infty \frac{\hbar^2 |\vb{k}|^2}{2 m}
     \matrixel{\mathrm{FS}}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}}}{\mathrm{FS}} \dd{\vb{k}}
-    \\
+\end{aligned}$$
+
+The matrix element
+$$\matrixel{\mathrm{FS}}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}}}{\mathrm{FS}} \dd{\vb{k}}$$
+is either $$0$$ or $$1$$, depending on whether $$\vb{k}$$
+is outside the Fermi sphere, or, equivalently,
+whether $$|\vb{k}|$$ is above or below $$k_F$$.
+We can write this fact by introducing a
+[Heaviside step function](/know/concept/heaviside-step-function/) $$\Theta(k)$$:
+
+$$\begin{aligned}
+    E^{(0)}
     &= \frac{\hbar^2 V}{8 \pi^3 m} \int_{-\infty}^\infty |\vb{k}|^2 \: \Theta(k_F - |\vb{k}|) \dd{\vb{k}}
 \end{aligned}$$
 
-In spherical coordinates,
-we evaluate the integral and find that $$E^{(0)}$$ is proportional to $$k_F^5$$:
+We evaluate this in
+[spherical coordinates](/know/concept/spherical-coordinates/)
+and find that $$E^{(0)}$$ is proportional to $$k_F^5$$:
 
 $$\begin{aligned}
     E^{(0)}
     &= \frac{\hbar^2 V}{8 \pi^3 m} \int_0^{2 \pi}
     \int_0^\pi \int_0^\infty \Big( |\vb{k}|^2 \: \Theta(k_F - |\vb{k}|) \Big) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi}
     \\
-    &= \frac{\hbar^2 V}{8 \pi^3 m} 4 \pi \int_0^{k_F} |\vb{k}|^4 \dd{|\vb{k}|}
-    = \frac{\hbar^2 V}{2 \pi^2 m} \bigg[ \frac{|\vb{k}|^5}{5} \bigg]_0^{k_F}
-    = \frac{\hbar^2 V}{10 \pi^2 m} k_F^5
+    &= \frac{\hbar^2 V}{8 \pi^3 m} \: 4 \pi \int_0^{k_F} |\vb{k}|^4 \dd{|\vb{k}|}
+    \\
+    &= \frac{\hbar^2 V}{10 \pi^2 m} k_F^5
 \end{aligned}$$
 
 In general, it is more useful to consider
 the average kinetic energy per electron $$E^{(0)} / N$$,
-which we find to be as follows, using that $$k_F^3 = 3 \pi^2 n$$:
+which we find to be as follows,
+using that $$k_F^3 = 3 \pi^2 N / V$$:
 
 $$\begin{aligned}
     \boxed{
         \frac{E^{(0)}}{N}
         = \frac{3 \hbar^2}{10 m} k_F^2
-        = \frac{3}{5} \epsilon_F
     }
-    \:\sim\: n^{2/3}
+    \:\:\propto\: n^{2/3}
 \end{aligned}$$
 
-Traditionally, this is expressed using a dimensionless parameter $$r_s$$,
+Traditionally, this is rewritten using the **Wigner-Seitz radius** $$r_s$$,
 defined as the radius of a sphere containing a single electron,
-measured in Bohr radii $$a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m)$$:
+measured in Bohr radii $$a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e_0^2 m)$$:
 
 $$\begin{aligned}
         \frac{4 \pi}{3} (a_0 r_s)^3
-        = \frac{1}{n}
-        = \frac{3 \pi^2}{k_F^3}
-        \quad \implies \quad
+        \equiv \frac{1}{n}
+        \qquad \implies \qquad
         r_s
-        = \Big( \frac{3}{4 \pi a_0^3 n} \Big)^{1/3}
-        = \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 k_F}
+        = \bigg( \frac{3}{4 \pi a_0^3 n} \bigg)^{1/3}
 \end{aligned}$$
 
-Such that the ground state energy can be rewritten in Rydberg units of energy like so:
+Note that this is dimensionless due to our choice of $$a_0$$ as a unit.
+In the Fermi gas, we have:
+
+$$\begin{aligned}
+        r_s
+        = \bigg( \frac{9 \pi}{4} \bigg)^{1/3} \frac{1}{a_0 k_F}
+        \qquad \implies \qquad
+        k_F
+        = \bigg( \frac{9 \pi}{4} \bigg)^{1/3} \frac{1}{a_0 r_s}
+\end{aligned}$$
+
+By inserting this into the ground state energy
+and using the definition of $$a_0$$, we can write:
 
 $$\begin{aligned}
     \frac{E^{(0)}}{N}
-    = \frac{3 \hbar^2}{10 m} \frac{4 \pi \varepsilon_0 e^2}{4 \pi \varepsilon_0 e^2} \frac{a_0^2 k_F^2}{a_0^2}
-    = \frac{3 e^2}{40 \pi \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{2/3} \frac{1}{a_0 r_s^2}
-    \approx \frac{2.21}{r_s^2} \; \mathrm{Ry}
+    &= \frac{3 \hbar^2}{10 m} \bigg( \frac{9 \pi}{4} \bigg)^{2/3} \frac{1}{a_0^2 r_s^2}
+    \\
+    &= \frac{3}{5} \bigg( \frac{9 \pi}{4} \bigg)^{2/3} \bigg( \frac{e_0^2}{8 \pi \varepsilon_0 a_0} \bigg) \frac{1}{r_s^2}
 \end{aligned}$$
 
+Where the last parenthesized expression is
+the Rydberg unit of energy $$\mathrm{Ry} \approx 13.6 \:\mathrm{eV}$$, so:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{E^{(0)}}{N}
+        \approx \frac{2.21}{r_s^2} \; \mathrm{Ry}
+    }
+\end{aligned}$$
+
+This result is found in a lot of literature.
+The choice of Rydberg units is simply a tradition.
+
+
 
-## With interactions
+## 1st order
 
-To include Coulomb interactions, let us try
-[time-independent pertubation theory](/know/concept/time-independent-perturbation-theory/).
-Clearly, this will give better results when the interaction is relatively weak, if ever.
+In the next term of the perturbation expansion,
+we start to include Coulomb interactions.
+Clearly, this will give better results when the interaction is relatively weak,
+but is that ever the case?
 
 The Coulomb potential is proportional to the inverse distance,
 and the average electron spacing is roughly $$n^{-1/3}$$,
-so the interaction energy $$E_\mathrm{int}$$ should scale as $$n^{1/3}$$.
-We already know that the kinetic energy $$E_\mathrm{kin} = E^{(0)}$$ scales as $$n^{2/3}$$,
-meaning perturbation theory should be reasonable
-if $$1 \gg E_\mathrm{int} / E_\mathrm{kin} \sim n^{-1/3}$$,
-so in the limit of high density $$n \to \infty$$.
+so the interaction energy $$E_\mathrm{int}$$ scales as $$n^{1/3}$$.
+We also know that the kinetic energy $$E_\mathrm{kin} = E^{(0)}$$
+is proportional to $$n^{2/3}$$,
+meaning that it is reasonable to use perturbation theory
+as long as $$1 \gg E_\mathrm{int} / E_\mathrm{kin} \propto n^{-1/3}$$,
+i.e. in the limit of high density $$n \to \infty$$.
 
 The two-body Coulomb interaction operator $$\hat{W}$$
 is as follows in second-quantized form:
 
 $$\begin{aligned}
     \hat{W}
-    = \frac{1}{2 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{e^2}{\varepsilon_0 |\vb{q}|^2}
+    = \frac{1}{2 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{e_0^2}{\varepsilon_0 |\vb{q}|^2}
     \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1}
 \end{aligned}$$
 
@@ -228,7 +199,7 @@ is then given by:
 $$\begin{aligned}
     E^{(1)}
     = \matrixel{\mathrm{FS}}{\hat{W}}{\mathrm{FS}}
-    = \frac{e^2}{2 \varepsilon_0 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
+    = \frac{e_0^2}{2 \varepsilon_0 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
     \matrixel{\mathrm{FS}}{
         \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1}
     }{\mathrm{FS}}
@@ -246,17 +217,17 @@ Let $$s = s_1$$ and $$\vb{k} = \vb{k}_1$$:
 
 $$\begin{aligned}
     E^{(1)}
-    &= \frac{e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
+    &= \frac{e_0^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
     \matrixel{\mathrm{FS}}{
         \hat{c}_{s, \vb{k} + \vb{q}}^\dagger \hat{c}_{s, \vb{k}}^\dagger \hat{c}_{s, \vb{k} + \vb{q}} \hat{c}_{s, \vb{k}}
     }{\mathrm{FS}}
     \\
-    &= \frac{- e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
+    &= \frac{- e_0^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
     \matrixel{\mathrm{FS}}{
         \big( \hat{c}_{s, \vb{k} + \vb{q}}^\dagger \hat{c}_{s, \vb{k} + \vb{q}}\big) \big(\hat{c}_{s, \vb{k}}^\dagger \hat{c}_{s, \vb{k}}\big)
     }{\mathrm{FS}}
     \\
-    &= \frac{- e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
+    &= \frac{- e_0^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
     \Theta(k_F - |\vb{k}|) \:\Theta(k_F - |\vb{k} \!+\! \vb{q}|)
 \end{aligned}$$
 
@@ -269,11 +240,11 @@ This yields the integration limit, and therefore leads to:
 
 $$\begin{aligned}
     E^{(1)}
-    &= \frac{- e^2}{(2 \pi)^3 \varepsilon_0} \sum_{\vb{k}}
+    &= \frac{- e_0^2}{(2 \pi)^3 \varepsilon_0} \sum_{\vb{k}}
     \int_0^{2 \pi} \!\!\int_0^\pi \!\!\int_0^\infty \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \frac{|\vb{q}|^2}{|\vb{q}|^2}
     \sin(\theta_q) \dd{|\vb{q}|} \dd{\theta_q} \dd{\varphi_q}
     \\
-    &= \frac{- e^2}{2 \pi^2 \varepsilon_0} \sum_{\vb{k}}
+    &= \frac{- e_0^2}{2 \pi^2 \varepsilon_0} \sum_{\vb{k}}
     \int_0^{2 k_F} \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \dd{|\vb{q}|}
 \end{aligned}$$
 
@@ -285,11 +256,11 @@ when we go to spherical coordinates $$(|\vb{k}|, \theta_k, \varphi_k)$$ for $$\v
 
 $$\begin{aligned}
     E^{(1)}
-    &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^\infty
+    &= \frac{- e_0^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^\infty
     \!\Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|)
     \: |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|}
     \\
-    &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^{k_F}
+    &= \frac{- e_0^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^{k_F}
     \!\Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|)
     \: |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|}
 \end{aligned}$$
@@ -335,13 +306,13 @@ substituting $$\xi \equiv \cos(\theta_k)$$:
 
 $$\begin{aligned}
     E^{(1)}
-    &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} 2 \int_0^{2 k_F} \!\!\!\int_0^{2 \pi} \!\!\int_0^{\arccos{|\vb{q}| / (2 k_F)}}
+    &= \frac{- e_0^2 V}{16 \pi^5 \varepsilon_0} 2 \int_0^{2 k_F} \!\!\!\int_0^{2 \pi} \!\!\int_0^{\arccos{|\vb{q}| / (2 k_F)}}
     \!\!\int_{|\vb{q}|/(2 \cos{\theta_k})}^{k_F} |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|}
     \\
-    &= \frac{e^2 V}{8 \pi^5 \varepsilon_0} 2 \pi \int_0^{2 k_F} \!\!\!\int_1^{|\vb{q}| / (2 k_F)}
+    &= \frac{e_0^2 V}{8 \pi^5 \varepsilon_0} 2 \pi \int_0^{2 k_F} \!\!\!\int_1^{|\vb{q}| / (2 k_F)}
     \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \frac{\sin(\theta_k)}{\sin(\theta_k)} \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|}
     \\
-    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
+    &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
     \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|}
 \end{aligned}$$
 
@@ -350,23 +321,23 @@ Evaluating these integrals:
 
 $$\begin{aligned}
     E^{(1)}
-    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
+    &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
     \bigg[ \frac{|\vb{k}|^3}{3} \bigg]_{|\vb{q}|/(2 \xi)}^{k_F} \dd{\xi} \dd{|\vb{q}|}
     \\
-    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
+    &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
     \bigg( \frac{k_F^3}{3} - \frac{|\vb{q}|^3}{24 \xi^3} \bigg) \dd{\xi} \dd{|\vb{q}|}
     \\
-    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
+    &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
     \bigg[ \frac{k_F^3}{3} x + \frac{|\vb{q}|^3}{48 \xi^2} \bigg]_{|\vb{q}| / (2 k_F)}^1 \dd{|\vb{q}|}
     \\
-    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
+    &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
     \bigg( \frac{k_F^3}{3} + \frac{|\vb{q}|^3}{48} - \frac{k_F^2 |\vb{q}|}{4} \bigg) \dd{|\vb{q}|}
     \\
-    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \bigg[ \frac{k_F^3 |\vb{q}|}{3} + \frac{|\vb{q}|^4}{192} - \frac{k_F^2 |\vb{q}|^2}{8} \bigg]_0^{2 k_F}
+    &= \frac{- e_0^2 V}{4 \pi^4 \varepsilon_0} \bigg[ \frac{k_F^3 |\vb{q}|}{3} + \frac{|\vb{q}|^4}{192} - \frac{k_F^2 |\vb{q}|^2}{8} \bigg]_0^{2 k_F}
     \\
-    &= \frac{- e^2 V}{16 \pi^4 \varepsilon_0} k_F^4
-    = \frac{- e^2 N}{16 \pi^4 \varepsilon_0 n} k_F^4
-    = -\frac{3 e^2 N}{16 \pi^2 \varepsilon_0} k_F
+    &= \frac{- e_0^2 V}{16 \pi^4 \varepsilon_0} k_F^4
+    = \frac{- e_0^2 N}{16 \pi^4 \varepsilon_0 n} k_F^4
+    = -\frac{3 e_0^2 N}{16 \pi^2 \varepsilon_0} k_F
 \end{aligned}$$
 
 Per particle, the first-order energy correction $$E^{(1)}$$
@@ -375,7 +346,7 @@ is therefore found to be as follows:
 $$\begin{aligned}
     \boxed{
         \frac{E^{(1)}}{N}
-        = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} k_F
+        = -\frac{3 e_0^2}{16 \pi^2 \varepsilon_0} k_F
     }
 \end{aligned}$$
 
@@ -383,8 +354,8 @@ This can also be written using the parameter $$r_s$$ introduced above, leading t
 
 $$\begin{aligned}
     \frac{E^{(1)}}{N}
-    = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \frac{a_0 k_F}{a_0}
-    = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 r_s}
+    = -\frac{3 e_0^2}{16 \pi^2 \varepsilon_0} \frac{a_0 k_F}{a_0}
+    = -\frac{3 e_0^2}{16 \pi^2 \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 r_s}
 \end{aligned}$$
 
 Consequently, for sufficiently high densities $$n$$,